#
Problem
of the Week

Problem
D and Solution

Halfway
to the Other Side

## Problem

Cube \(PQRSTUVW\) has side length
\(2\). Point \(M\) is the midpoint of edge \(UT\). Determine the area of \(\triangle MQR\).

## Solution

We first draw \(RT\).

In \(\triangle RWT\), \(\angle RWT = 90^\circ\) and \(RW = WT = 2\).

By the Pythagorean Theorem in \(\triangle
RWT\), \(RT^2= RW^2 + WT^2= 2^2 +
2^2=8\).

Therefore, \(RT = \sqrt{8}\), since
\(RT>0\).

\(\triangle MQR\) has base equal to
the length of \(QR\), which is \(2\).

Notice that the height of \(\triangle
MQR\) is equal to the distance from side \(QR\) of the cube to side \(UT\) of the cube, which is equal to the
length of \(RT\) or \(\sqrt{8}\).

Therefore, area of \(\triangle MQR =
\dfrac{\text{base}\times \text{height}}{2} = \dfrac{2 \times
\sqrt{8}}{2}= \sqrt{8}\) units squared.